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Those Fascinating Numbers
Page7(27 of 451)

Those Fascinating Numbers 7 21 • the smallest integer 1 whose sum of divisors is a fifth power: here σ(21) = 25; • the smallest 2-hyperperfect number: a number n is said to be 2-hyperperfect if it can be written as n = 1 + 2 d|n 1dn d, which is equivalent to the condition 2σ(n) = 3n + 1; the sequence of numbers satisfying this property begins as follows: 21, 2 133, 19 521, 176 661, 129 127 041, . . . ; more generally, a number n is said to be hyperperfect if there exists a positive integer k such that n = 1 + k d|n 1dn d, (1) in which case we also say that n is k-hyperperfect 9; the following table contains some k-hyperperfect numbers along with their factorization: 9A 1-hyperperfect number is simply a perfect number. lt is easy to show that relation (1) is equivalent to kσ(n) = (k + 1)n + (k − 1). (2) Also, it is clear that a prime power pα, with α ≥ 1, cannot be hyperperfect. Furthermore, it follows immediately from (1) that if n is k-hyperperfect, then n ≡ 1 (mod k) and moreover that σ(n) = n + 1 + n − 1 k . (3) This last relation proves to be an excellent tool to determine if a given integer n is a hyperperfect number and also to construct, using a computer, a list of hyperperfect numbers. Indeed, it follows from (3) that n is a hyperperfect number ⇐⇒ n − 1 σ(n) − n − 1 is an integer. It also follows from (3) that the smallest prime factor of such an integer n is larger than k. Indeed, assume that p|n with p ≤ k. We would then have that n/p is a proper divisor of n, in which case σ(n) n + 1 + n p ≥ n + 1 + n k n + 1 + n − 1 k = σ(n), a contradiction. It follows from this that a hyperperfect number which is not perfect is odd. On the other hand, if n is a square-free k-hyperperfect number, then k must be even. Assume the contrary, that is that k is odd. As we just saw, n must be odd, unless k = 1, in which case n would be perfect and even. But then we would have that n = 2p−1(2p − 1) for a certain prime number p ≥ 3, in which case n would not be square-free. We therefore have that n is odd. Now, because of (2), we have kσ(n) = 2 k + 1 2 n + k − 1 2 . (4) If k ≡ 1 (mod 4), then it follows from (4) that kσ(n) = 2 (odd + even) = 2 × odd, while if k ≡ 3 (mod 4), then kσ(n) = 2 (even + odd) = 2 × odd, which means that 2 σ(n), in which case n is prime, since n is square-free.